§ Open mapping theorem

• Given a surjective continuous linear map $f: X \to Y$, image of open unit ball is open.
• Immediate corollary: image of open set is open (translate/scale open unit ball around by linearity) to cover any open set with nbhds.

§ Quick intuition

• Intuition 1: If the map $f$ we bijective, then thm is reasonably believeable given bounded/continuous inverse theorem, since $f^{-1}$ would be continuous, and thus would map open sets to open sets, which would mean that $f$ does the same.
• In more detail: suppose $f^{-1}$ exists and is continuous. Then $f(U) = V$implies $(f^{-1})^{-1}(U) = V$. Since $f^{-1}$ is continuous, the iverse image of an open set ( $U$) is open, and thus $V$ is open.

§ Why surjective

• Consider the embedding $f : x \mapsto (x, x)$ from $\mathbb R$ to $\mathbb R^2$.
• The full space $\R$ is open in the domain of $f$, but is not open in $\mathbb R^2$, since any epsilon ball around any point in the diagonal $(x, x)$ would leak out of the diagonal.
• Thus, not every continuous linear map maps open sets to open sets.

§ Proof

• TODO